`|{:(x,4, y+z),(y,4,z+x),(z,4,x+y):}|` is equal to:

Options

A. (a) `4`
B. (b) `4+y+z`
C. (c) `xyz`
D. (d) `0`

Answer

To solve the determinant \( | \begin{vmatrix} x & 4 & y+z \\ y & 4 & z+x \\ z & 4 & x+y \end{vmatrix} | \), we will follow these steps:

### Step 1: Write down the determinant
We start with the determinant:
\[ \Delta = \begin{vmatrix} x & 4 & y+z \\ y & 4 & z+x \\ z & 4 & x+y \end{vmatrix} \]

### Step 2: Perform column operations
We will add the third column to the first column. This gives us:
\[ \Delta = \begin{vmatrix} x + (y+z) & 4 & y+z \\ y + (z+x) & 4 & z+x \\ z + (x+y) & 4 & x+y \end{vmatrix} \]
This simplifies to:
\[ \Delta = \begin{vmatrix} x+y+z & 4 & y+z \\ y+z+x & 4 & z+x \\ z+x+y & 4 & x+y \end{vmatrix} \]

### Step 3: Factor out common terms
Now, we can factor out \( (x+y+z) \) from the first column and \( 4 \) from the second column:
\[ \Delta = (x+y+z) \cdot 4 \cdot \begin{vmatrix} 1 & 1 & y+z \\ 1 & 1 & z+x \\ 1 & 1 & x+y \end{vmatrix} \]

### Step 4: Analyze the new determinant
The new determinant is:
\[ \begin{vmatrix} 1 & 1 & y+z \\ 1 & 1 & z+x \\ 1 & 1 & x+y \end{vmatrix} \]
Notice that the first two columns are identical, which implies that the determinant is zero:
\[ \begin{vmatrix} 1 & 1 & y+z \\ 1 & 1 & z+x \\ 1 & 1 & x+y \end{vmatrix} = 0 \]

### Step 5: Conclude the value of the original determinant
Thus, we have:
\[ \Delta = (x+y+z) \cdot 4 \cdot 0 = 0 \]

### Final Answer
The value of the determinant is:
\[ \Delta = 0 \]